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Question Number 70598 by Shamim last updated on 06/Oct/19

If a,b,c ∈ ℜ and (a^2 /(b+c))+(b^2 /(c+a))+(c^2 /(a+b))=((12)/(a+b+c)) , (a/(b+c))+(b/(a+c))+(c/(a+b))=(4/3) then find a+b+c=?

$$\mathrm{If}\:\mathrm{a},\mathrm{b},\mathrm{c}\:\in\:\Re\:\mathrm{and}\:\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{b}+\mathrm{c}}+\frac{\mathrm{b}^{\mathrm{2}} }{\mathrm{c}+\mathrm{a}}+\frac{\mathrm{c}^{\mathrm{2}} }{\mathrm{a}+\mathrm{b}}=\frac{\mathrm{12}}{\mathrm{a}+\mathrm{b}+\mathrm{c}} \\ $$$$,\:\frac{\mathrm{a}}{\mathrm{b}+\mathrm{c}}+\frac{\mathrm{b}}{\mathrm{a}+\mathrm{c}}+\frac{\mathrm{c}}{\mathrm{a}+\mathrm{b}}=\frac{\mathrm{4}}{\mathrm{3}}\:\mathrm{then}\:\mathrm{find}\:\mathrm{a}+\mathrm{b}+\mathrm{c}=? \\ $$

Answered by som(math1967) last updated on 06/Oct/19

(a^2 /(b+c)) +a+(b^2 /(c+a)) +b +(c^2 /(c+a)) +c−(a+b+c) =((12)/(a+b+c)) ((a(a+b+c))/(b+c)) +((b(a+b+c))/(c+a))+((c(a+b+c))/(a+b)) −(a+b+c)=((12)/(a+b+c)) (a+b+c)((a/(b+c)) +(b/(c+a)) +(c/(a+b)) −1)=((12)/(a+b+c)) (a+b+c)((4/3) −1)=((12)/(a+b+c)) ★ (((a+b+c)^2 )/3)=12 (a+b+c)^2 =36 ⇒(a+b+c)=±6 ★(a/(b+c)) +(b/(c+a)) +(c/(a+b)) =(4/3)

$$\frac{{a}^{\mathrm{2}} }{{b}+{c}}\:+{a}+\frac{{b}^{\mathrm{2}} }{{c}+{a}}\:+{b}\:+\frac{{c}^{\mathrm{2}} }{{c}+{a}}\:+{c}−\left({a}+{b}+{c}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{12}}{{a}+{b}+{c}} \\ $$$$\frac{{a}\left({a}+{b}+{c}\right)}{{b}+{c}}\:+\frac{{b}\left({a}+{b}+{c}\right)}{{c}+{a}}+\frac{{c}\left({a}+{b}+{c}\right)}{{a}+{b}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\left({a}+{b}+{c}\right)=\frac{\mathrm{12}}{{a}+{b}+{c}} \\ $$$$\left({a}+{b}+{c}\right)\left(\frac{{a}}{{b}+{c}}\:+\frac{{b}}{{c}+{a}}\:+\frac{{c}}{{a}+{b}}\:−\mathrm{1}\right)=\frac{\mathrm{12}}{{a}+{b}+{c}} \\ $$$$\left({a}+{b}+{c}\right)\left(\frac{\mathrm{4}}{\mathrm{3}}\:−\mathrm{1}\right)=\frac{\mathrm{12}}{{a}+{b}+{c}}\:\:\:\bigstar \\ $$$$\frac{\left({a}+{b}+{c}\right)^{\mathrm{2}} }{\mathrm{3}}=\mathrm{12} \\ $$$$\left({a}+{b}+{c}\right)^{\mathrm{2}} =\mathrm{36}\:\Rightarrow\left({a}+{b}+{c}\right)=\pm\mathrm{6} \\ $$$$\bigstar\frac{{a}}{{b}+{c}}\:+\frac{{b}}{{c}+{a}}\:+\frac{{c}}{{a}+{b}}\:=\frac{\mathrm{4}}{\mathrm{3}} \\ $$

Commented by Shamim last updated on 06/Oct/19

tnks

$$\mathrm{tnks} \\ $$

Commented by Rasheed.Sindhi last updated on 06/Oct/19

∨ ⊓∣⊏∈!

$$\vee\:\:\sqcap\shortmid\sqsubset\in! \\ $$

Commented by som(math1967) last updated on 06/Oct/19

Nice use of mathematical symbols

$${Nice}\:{use}\:{of}\:{mathematical}\:{symbols} \\ $$

Commented by peter frank last updated on 06/Oct/19

good

$${good} \\ $$

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